A261342 -id:A261342 - cp's OEIS frontend

A261205 Numbers k such that floor(k^(1/m)) divides k for all integers m >= 1.

1, 2, 3, 4, 6, 8, 12, 16, 20, 24, 30, 36, 42, 48, 64, 72, 80, 120, 210, 240, 288, 324, 420, 528, 552, 576, 600, 624, 900, 1260, 1764, 1848, 1980, 3024, 6480, 8100, 8280, 11880, 14160, 14280, 14400, 14520, 14640, 28560, 43680, 44520, 46872, 50400, 175560, 331200, 346920, 491400, 809100, 3418800, 4772040, 38937600, 203918400, 2000862360

Offset: 1

Yan A. Denenberg and Max Alekseyev, Aug 11 2015

Is this a finite sequence?

There are no other terms below 10^23. - Giovanni Resta, Aug 13 2015

			From _Michel Marcus_, Aug 13 2015: (Start)
For k=1 to 9, we have the following floored roots:
  k=1: 1, 1, ...
  k=2: 2, 1, 1, ...
  k=3: 3, 1, 1, ...
  k=4: 4, 2, 1, 1, ...
  k=5: 5, 2, 1, 1, ...
  k=6: 6, 2, 1, 1, ...
  k=7: 7, 2, 1, 1, ...
  k=8: 8, 2, 2, 1, 1, ...
  k=9: 9, 3, 2, 1, 1, ...
where one can see that 5, 7 and 9 are not terms. (End)

Cf. A261206, A261341, A261342.

Subsequence of A006446.

fQ[n_] := Block[{d, k = 2, lst = {}}, While[d = Floor[n^(1/k)]; d > 1, AppendTo[lst, d]; k++]; Union[ IntegerQ@# & /@ (n/Union[lst])] == {True}]; k = 4; lst = {1, 2, 3}; While[k < 10^6, If[fQ@ k, AppendTo[lst, k]; Print@ k]; k++]; lst (* Robert G. Wilson v, Aug 15 2015 *)

is(n) = my(k,t); k=2; while( (t=sqrtnint(n, k)) > 1, if(n%t, return(0)); k++); 1
n=1; while(n<10^5,if(is(n),print1(n,", "));n++) /* Able to generate terms < 10^5 */ \\ Derek Orr, Aug 12 2015

A261206 Numbers j such that ceiling(j^(1/k)) divides j for all integers k >= 1.

Original entry on oeis.org

1, 2, 4, 6, 12, 36, 132, 144, 156, 900, 3600, 4032, 7140, 18360, 44100, 46440, 4062240, 9147600, 999999000000

Offset: 1

Max Alekseyev, Aug 11 2015

Is this a finite sequence?
It is possible to generalize this class of sequences by taking some integer-valued function f(j,k) decreasing in k such that f(j,1) = j and f(j,m) = c (for example, c=1 or c=2) for all sufficiently large m and considering those j that are divisible by all of f(j,1), f(j,2), ... If f(j,k) is slowly decreasing in k, then the set of corresponding j's is likely to have a very small number (if any) of terms, while if f(j,k) decreases rapidly, then there will be too many suitable j's. I believe the balance is achieved at functions like f(j,k) = floor(j^(1/k)) so that f(j,k) stabilizes to c at k ~= log(j). - Max Alekseyev, Aug 16 2015
If it exists, a(20) > 10^35. - Jon E. Schoenfield, Oct 17 2015

Cf. A261205, A261341, A261342.

Subsequence of all of A087811, A002620, A261011, A261417.

is(n) = my(k,t); if(n==1,return(1)); if(n%2,return(0)); k=2; while( (t=ceil((n-.5)^(1/k)))>2, if(n%t,return(0)); k++); 1
n=1;while(n<10^5,if(is(n),print1(n,", "));n++) /* Able to generate terms < 10^5 */ \\ Derek Orr, Aug 12 2015

A261341 Numbers n such that round(n^(1/k)) divides n for all integers k>=1.

Original entry on oeis.org

1, 2, 4, 6, 12, 30, 36, 42, 72, 240, 420, 600, 900, 1560, 1764, 3600, 6084, 8100, 46440, 1742400, 4062240, 35814240

Offset: 1

Max Alekseyev, Aug 15 2015

There are no other terms below 10^16.

Is this a finite sequence?

Cf. A261205, A261206

Subsequence of A006446 and A261342.

isA[n_] :=
Block[{t},
  For[k = 2, (t = Floor[1/2 + n^(1/k)]) >= 2, k++,
   If[Mod[n, t] != 0, Return[False]]]; Return[True]]
Select[Range[1, 100000], isA[#] &] (* Julien Kluge, Apr 04 2016 *)

{ isA261341(n) = my(k,t); k=2; until(t<=2, t=round(sqrtn(n,k)); if(n%t,return(0)); k++); 1; }

cp's OEIS Frontend

A261205 Numbers k such that floor(k^(1/m)) divides k for all integers m >= 1.

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A261206 Numbers j such that ceiling(j^(1/k)) divides j for all integers k >= 1.

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A261341 Numbers n such that round(n^(1/k)) divides n for all integers k>=1.

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